Orbital velocities in the Schwartzschild geometry

[Altabeh]

You mean $$L=(1-2m/r)\dot{t}^2-r^2\dot{\phi}^2$$?
Congratulations, you have rediscovered post 2.




You rediscovered post 6. You are on the right track.

What about the rest and most important part? Hahaaa!


AB

 

[starthaus]

And it's not my fault that you can't recognize basic special relativity and difference between proper and coordinate and finally how playing with equations would lead to a completely algebraically proven "formula"; not arrangement of terms and leaving $$dr/ds$$ unidentified so as to call the equation "general"! Haha!

If your complaint about my derivations is that I call $$\frac{d^2r}{d\tau^2}$$ proper acceleration, then so be it. I can live with that.

 

[starthaus]

What about the rest and most important part? Hahaaa!AB

Yes, you managed to get the same results I got several days before you. Good job.

 

[Altabeh]

If your complaint about my derivations is that I call $$\frac{d^2r}{d\tau^2}$$ proper acceleration, then so be it. I can live with that.

Come on! You know that it's wrong and living with wrong will make you believe in wrong, of course! You can't get away with it!

Yes, you managed to get the same results I got several days before you. Good job.

I don't see the result I gave earlier in this page in any of your posts so don't attempt to attach my "formula" to your nonsense equations. Period.

AB

 

[starthaus]

I don't see the result I gave earlier in this page in any of your posts so don't attempt to attach my "formula" to your nonsense equations. Period.

AB

Not my problem that you can only read your own posts.

 

[yuiop]

I was certain all along that dr/dt=0 does not imply that d^2r/dt^2=0 but I did not know how to prove it until I came across this article http://mathforum.org/library/drmath/view/65095.html in Dr math. Seeing as how Starhaus has not acknowledged his mistake I will assume he still does not get it and will elaborate on it.

f(x) = x^2
f'(x) = 2x
f''(x) = 2

When x=0:

f(x=0) = 0
f'(x=0) = 0
f''(x=0) = 2 ... Ta da! ... Non-zero!

Quite a simple proof. You should be able to get it now surely?

 

[starthaus]

I was certain all along that dr/dt=0 does not imply that d^2r/dt^2=0 but I did not know how to prove it until I came across this article http://mathforum.org/library/drmath/view/65095.html in Dr math. Seeing as how Starhaus has not acknowledged his mistake I will assume he still does not get it and will elaborate on it.

I get it all right. It is you who doesn't.

f(x) = x^2
f'(x) = 2x
f''(x) = 2

When x=0:

f(x=0) = 0
f'(x=0) = 0
f''(x=0) = 2 ... Ta da! ... Non-zero!

Quite a simple proof. You should be able to get it now surely?

You still don't understand the difference between a function and the value of a function in a point. So, no "when x=0" applies. Take that calculus class, it will do you a lot of good.

 

[yuiop]

You still don't understand the difference between a function and the value of a function in a point. So, no "when x=0" applies. Take that calculus class, it will do you a lot of good.

You don't have the flexibility to realize that if a function f(x) is valid for all x then one of the values that x can take is zero. Any conclusions drawn from assuming x=0 are only valid at that point in the curve and for most people here it is not a problem as long as the context is understood. That means you have to read some of the surrounding text rather than just focusing on the mathematical symbols and being too rigid in your formalisms. Even when I demostrated to you that your mathematics leads to the conclusion that a ball thrown up in the air stops at the apogee and does not fall back down, you would rather consider that there was something wrong with nature before you would consider that you made a mistake with your mathematics. Here is a hot tip. If your mathematics do not agree with what is seen with nature, it is the mathematics that is probably wrong.

 

[yuiop]

Nice work in post #132 Altabeh. I will bookmark it and come back to it. I am sure I can learn some new things from your material.

Thanks.

 

[starthaus]

You don't have the flexibility to realize that if a function f(x) is valid for all x then one of the values that x can take is zero. Any conclusions drawn from assuming x=0 are only valid at that point in the curve and for most people here it is not a problem as long as the context is understood. That means you have to read some of the surrounding text rather than just focusing on the mathematical symbols and being too rigid in your formalisms.

I don't understand why you insist in continuing to embarass yourself by showing your ignorance in terms of calculus. Especially since you have been shown several derivations that do not employ the $$dr=0$$ hack.

Even when I demostrated to you that your mathematics leads to the conclusion that a ball thrown up in the air stops at the apogee and does not fall back down, you would rather consider that there was something wrong with nature before you would consider that you made a mistake with your mathematics.

While you "demonstration" is correct, it is totally irrelevant. It simply illustrates your inability to tell the difference between a function and its value in a point. You could easily remedy this if you took a class in calculus. You can't really pretend that you're doing physics when you fail basic calculus. It is really simple, kev, calculus 101 teaches you that if $$f(x)=constant$$ then
$$\frac{df}{dx}=0$$ for all x. There is no way around it. Can you generalize this to the case$$\frac{df}{dx}=constant$$ implies $$\frac{d^2f}{dx^2}=0$$ for all x?

 

[yuiop]

I don't understand why you insist in continuing to embarass yourself by showing your ignorance in terms of calculus. Especially since you have been shown several derivations that do not employ the $$dr=0$$ hack.

What is even more embarassing is that even with my limited knowledge of calculus (which is improving all the time) I can out perform you in terms of arriving at correct solutions. I have done several derivations in this thread (and other threads) with and without the so called dr=0 "hack". Unlike you I have the flexibilty to work with either method and understand the domain of validity of each method.

While you "demonstartion" is correct, it simply illustrates your inability to tell the difference between a function and its value in a point. You could easily remedy this if you took a class in calculus. You can't really pretend that you're doing physics when you fail basic calculus. It is really simple, kev, calculus 101 teaches you that if $$f(x)=constant$$ then
$$\frac{df}{dx}=0$$ for all x. There is no way around it.

This is a nice attempt to create your own straw man argument here, but it is simply a distraction from the simple fact that your assertion that the acceleration of a particle at its apogee is zero, because its velocity is zero, is simply wrong.

I can tell the difference between a function and its value at a point and can apply either with equal ease. You on the other hand are unable to answer a simple question like what is the acceleration of a particle at its apogee, because as far as you are concerned there is no such thing as "when dr=0".

Although calculus is important for doing physics and I am working on improving my calculus abilities, I think a basic understanding of physics and algebra is an even more fundamental prerequisite and you seem to lack these.

 

[George Jones]

A reminder of a part of the Physics Forums Rules:

Guidelines on Langauge and Attitude:
Foul or hostile language will not be tolerated on Physics Forums. This includes profanity, obscenity, or obvious indecent language; direct personal attacks or insults; snide remarks or phrases that appear to be an attempt to "put down" another member; and other indirect attacks on a member's character or motives.

Please treat all members with respect, even if you do not agree with them. If you feel that you have been attacked, and the moderators or mentors have not yet gotten around to doing something about it, please report it using the "Report" button. If you choose to post a response, address only the substantive content, constructively, and ignore any personal remarks.

It is better to walk away from a possible confontation and come back later with constructive arguments.

 

[yuiop]

calculus 101 teaches you that if $$f(x)=constant$$ then
$$\frac{df}{dx}=0$$ for all x. There is no way around it. Can you generalize this to the case$$\frac{df}{dx}=constant$$ implies $$\frac{d^2f}{dx^2}=0$$ for all x?

Let us say we have a function of x such that f(x)= x^2.

When x=0 then f(x) = 0.
When x=1 then f(x) = 1.
When x=2 then f(x) = 4 and so on..

Now the first statement (When x=0 then f(x)=0) does not imply (f(x) = constant) and nor does it imply that (f(x) = 0 for all x). So when I say at the apogee x=0 and f(x)=0, I am not saying f(x)=0 for all x. You have to understand the context. Sometimes people use notation like (When x=0 then f(x=0) = 0) to make it clearer that (f(x)=0 for x=0) does not imply (f(x)=0 for all x) and maybe this is what is confusing you.

 

[starthaus]

This is a nice attempt to create your own straw man argument here, but it is simply a distraction from the simple fact that your assertion that the acceleration of a particle at its apogee is zero, because its velocity is zero, is simply wrong.

This is not what I've been telling you. I've been showing you that you don't understand the implication $$f(x)=const$$ => $$\frac{df}{dx}=0$$ in order to help you undestand why you can't just plug in $$dr=0$$ in the geodesic equation for radial motion.

I can tell the difference between a function and its value at a point and can apply either with equal ease.

The fact that you persist demonstrates that you still can't.

 

[starthaus]

Let us say we have a function of x such that f(x)= x^2.

That's a bad start. What you have is $$f(x)=\frac{dg}{dx}=0$$. What can you infer about $$\frac{df}{dx}$$?

 

[yuiop]

That's a bad start. What you have is $$f(x)=\frac{dg}{dx}=0$$. What can you infer about $$\frac{df}{dx}$$?

What I had was $$f(x)=x^2$$.

This infers $$\frac{df}{dx} = 2x$$

Which in turn infers $$\frac{d^2f}{dx^2} = 2$$

When x=0,

$$f(x)=0$$

$$\frac{df}{dx} = 0$$

$$\frac{d^2f}{dx^2} = 2$$

 

[starthaus]

What I had was $$f(x)=x^2$$.

What you have is $$f(x)=constant=0$$

As in the case of radial motion, you are not allowed to hack in $$\frac{dr}{ds}=0$$ in the equation of motion:

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 = 0$$

If you do, you'll get the wrong result.
You are allowed to set terms to 0 in the metric (indicating that there is no motion in the respective direction) but you are not allowed to set terms to 0 in the differential equation that describes the motion. Because if you hack in terms like $$\frac{dr}{ds}=0$$ in the equation of motion you end up with the wrong equation of motion.

 

[Altabeh]

I was certain all along that dr/dt=0 does not imply that d^2r/dt^2=0 but I did not know how to prove it until I came across this article http://mathforum.org/library/drmath/view/65095.html in Dr math. Seeing as how Starhaus has not acknowledged his mistake I will assume he still does not get it and will elaborate on it.

f(x) = x^2
f'(x) = 2x
f''(x) = 2

When x=0:

f(x=0) = 0
f'(x=0) = 0
f''(x=0) = 2 ... Ta da! ... Non-zero!

Quite a simple proof. You should be able to get it now surely?

Obviously mathematically correct and has nothing to do with nonsense claims like since it's the value at a point so it doesn't apply in general. Because one can find points at which the first derivative of a function iz zero while the second isn't, the claim $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is not generally correct.

This one another time shows how limited the guy's knowledge of basic algebra and calculus is.

AB

 

[Altabeh]

What I had was $$f(x)=x^2$$.

This infers $$\frac{df}{dx} = 2x$$

Which in turn infers $$\frac{d^2f}{dx^2} = 2$$

When x=0,

$$f(x)=0$$

$$\frac{df}{dx} = 0$$

$$\frac{d^2f}{dx^2} = 2$$

The "Second Derivative" method we use to find extrema of a given function says that

Step A. Let f(x) be a differentiable function on a given interval and let f'' be continuous at stationary point. Find f'(x) and solve the equation f'(x) = 0 given. Let x = a, b, ... be solutions.

Step B. Case (i) : If f''(a) < 0 then f(a) is maximum. Case (ii): If f''a)> 0 then f(a) is minimum.

This basic material is way below the level of discussion, but since some people sound quite unfamiliar to it, was recalled here.

AB

 

[yuiop]

If you calculate the lagrangian expression carefully, you should be getting exactly


$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

i.e. the same thing as the geodesic expression.

If the solution using the Langragian method is supposed to be the same as the that obtained from the geodesic method, then yes, this is what you should be getting if you calculate the Langragian expression carefully. This is what Espen got from the geodesic mthod. It is not what you obtained from Langragian. What you obtained in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" was:

$$-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0$$

which evaluates to:

$$\alpha/2*d/ds(1/\alpha*2dr/ds) + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

which is definitely not the same as the solution obtained by Espen. Is this your way of admitting that you made a mistake and did not evaluate the lagrangian method carefully?

Hint:

$$\frac{d}{ds}(\frac{2/\alpha}{dr/ds})=\frac{2}{\alpha^2}(\alpha\frac{d^2r}{ds^2}-(dr/ds)^2\frac{d\alpha}{dr})$$

The term $$\frac{d}{ds}(\frac{2/\alpha}{dr/ds})$$

does not appear in your expression derived from the langragian.

Hint:

$$\frac{d}{ds}(\frac{2/\alpha}{dr/ds}) \ne \frac{d}{ds}(\frac{1}{\alpha}*\frac{2dr}{ds})$$

 

[starthaus]

If the solution using the Langragian method is supposed to be the same as the that obtained from the geodesic method,

It isn't "supposed", it IS.

then yes, this is what you should be getting if you calculate the Langragian expression carefully. This is what Espen got from the geodesic mthod. It is not what you obtained from Langragian.

...because you did not finish the calculations. Use the hint I gave you. at post 136. If you calculate the derivatives correctly you will find out that , contrary to your claims, the lagrangian method has produced the same exact equation of motion as the geodesic method, as it should.

 

[starthaus]

Obviously mathematically correct and has nothing to do with nonsense claims like since it's the value at a point so it doesn't apply in general. Because one can find points at which the first derivative of a function iz zero while the second isn't, the claim $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is not generally correct.

Let's try again:
$$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.
Since you have been given that $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?

This one another time shows how limited the guy's knowledge of basic algebra and calculus is.

AB

You miss the point completely. Read post 157 then try comparing the solutions to the two ODE's shown here:

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

 

[espen180]

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

You do realize what the equation means? It relates the values of the functions for different values of a geodesic parameter.

As a reminder, a geodesic is a curve $$g\left(t(p),r(p),\theta(p),\phi(p)\right)=g_0\left(s(p),0,0,0\right)$$, where $$p$$ is the parameter of the curve.

Assuming there is such an r (or value of the geodesic parameter) where dr/ds is zero (an example of such an r is the apogee radius, or the point of release from rest), at that r, equations 1 and 2 degenerate into each other. This is Algebra II tops.

 

[starthaus]

or the point of release from rest), at that r, equations 1 and 2 degenerate into each other. .

...but they DO NOT degenerate into each other ANYWHERE else over the WHOLE domain of definition of the function $$r=r(s)$$. This is the whole point as to why you, Al68,kev should not be HACKING $$\frac{dr}{ds}=0$$ into the equation of motion. In the ODE, $$\frac{dr}{ds}$$ represents a FUNCTION, not a value, so you are not allowed to set it to 0 (or any other value) because you destroy the equation. Read post 157.

This is Algebra II tops

No, it is calculus 101. And, after all the explanations, you still get it wrong.

 

[espen180]

...but they DO NOT degenerate into each other ANYWHERE else over the WHOLE domain of definition of the function $$r=r(s)$$. This is the whole point as to why you, Al68,kev should not be HACKING $$\frac{dr}{ds}=0$$ into the equation of motion. In the ODE, $$\frac{dr}{ds}$$ represents a FUNCTION, not a value, so you are not allowed to set it to 0 (or any other value) because you destroy the equation. Read post 157.

You are wrong. Here is an example.

Look at these equations, in which I have only changed the derivative to the fuction itself:

1. $$f^{\prime\prime}(x)+af(x)+b=0$$

2. $$f^{\prime\prime}(x)+b=0$$

I argue that assuming there is such an x that f(x)=0, at that x the solutions degenerate and f''(x)=-b for both equations, not f''(x)=0 as you claim.

Solution of 1: $$f(x)=-\frac{b}{a}+c_1\sin\left(\sqrt{a}x\right)+c_2\cos\left(\sqrt{a}x\right)=-\frac{b}{a}+c_3\sin\left(\sqrt{a}x+c_4\right)$$

Solution of 2: $$-\frac{b}{2}x^2+c_1x+c_2$$

Double derivative of 1: $$f^{\prime\prime}=-c_3a\sin\left(\sqrt{a}x+c_4\right)$$

Double derivative of 2: $$f^{\prime\prime}=-b$$

When the solution to 1 is zero: $$x_0=\frac{\arcsin\left(\frac{b}{c_3a}\right)-c_4}{\sqrt{a}}$$

Plugging x₀ into double derivative of 1: $$-c_3a\sin\left(\sqrt{a}x_0+c_4\right)=-c_3a\sin\left(\arcsin\left(\frac{b}{c_3a}\right)\right)=-b$$

as expected.

 

[starthaus]

I argue that assuming there is such an x that f(x)=0, at that x the solutions degenerate and f''(x)=-b for both equations,

at that x...but NOWHERE ELSE in the domain of definition of f(x). This is the part that you, kev, Al68, Altabeh seem unable to grasp. You are being asked to find out the general equation of motion, your hack produces ONLY the acceleration at the point of release.

Exercise: using your hack try finding

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

 

[espen180]

at that x...but NOWHERE ELSE in the domain of definition of f(x). This is the part that you, kev, Al68, Altabeh seem unable to grasp. You are being asked to find out the general equation of motion, your hack produces ONLY the acceleration at the point of release.

The problem statement clearly spesified where the solution needed to be valid; at the apogee radius or drop point.

In a more general solution, such an assumption is not made, as can be seen in many posts here where work to reach such a solution has been/is being done.

 

[espen180]

Exercise: using your hack try finding

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

That's like asking what a photon's rest frame is like. You are trying to implement a model outside it's area of validity.

 

[starthaus]

The problem statement clearly spesified where the solution needed to be valid; at the apogee radius or drop point.

In what post did you specify this?
Even if your claim is true, what precluded you to understand what I have been telling you for tens of posts?

 

[starthaus]

That's like asking what a photon's rest frame is like.

It is good that you are starting to see the hack for what it is.

You are trying to implement a model outside it's area of validity.

The exercise I gave you is fully solvable with the tools I have given you, you simply need to stop trying to hack the ODE.

 

[Altabeh]

$$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.
Since you have been given that $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?

Nothing can be inferred about $$\frac{d^2r}{ds^2}$$ because kev gave you a very keen counter-example to the hack $$dr/ds=0\Rightarrow d^2r/ds^2=0$$. We don't know what form $$r$$ would have as a function of some (affine) parameter so that we are not allowed to generally start talking about all values of $$s$$ and that whether your hack works or not.

You miss the point completely. Read post 157 then try comparing the solutions to the two ODE's shown here:

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

You're the one who misses the point and doesn't listen to my notes. By now it's crystal clear that you're not even familiar with ODE's and boundary conditions. Actually a particle following a geodesic near any gravitating body would be considered momentarity at rest at any point* which means that along the geodesic $$dt/ds=0$$ for the particle. This never does mean that the particle is no longer going to have an acceleration but the object now has a uniform rest acceleration. If you have read a little about this, then you wouldn't dare to start giving us nonsense claims/equations. For example, you can read about this in this book "Relativity" By J. Rice. Purchase the book and don't waste our time by your hacks in calculus and algebra.

*For another usage in GR see for example "A first course in GR" by B. Schutz p. 254.

 

[starthaus]

Nothing can be inferred about $$\frac{d^2r}{ds^2}$$ because kev gave you a very keen counter-example to the hack $$dr/ds=0\Rightarrow d^2r/ds^2=0$$.

Let's try again, since you have been given that the function $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?
Keep in mind that $$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.

We don't know what form $$r$$ would have as a function of some (affine) parameter so that we are not allowed to generally start talking about all values of $$s$$ and that whether your hack works or not.

Irrelevant.

You're the one who misses the point and doesn't listen to my notes. By now it's crystal clear that you're not even familiar with ODE's and boundary conditions.

This is not about boundary conditions. Since when do you plug boundary conditions straight into the ODE? LOL

Actually a particle following a geodesic near any gravitating body would be considered momentarity at rest at any point* which means that along the geodesic $$dt/ds=0$$ for the particle.

Irrelevant in finding the trajectory $$r=r(t)$$ or the velocity $$v=v(t)$$ or the acceleration $$a=a(t)$$.

If you think otherwise, using your hack, find these:

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

The hack is only good only for finding the acceleration at the release point. Worthless for anything else.

 

[espen180]

The hack is only good only for finding the acceleration at the release point. Worthless for anything else.

Again, that is everything it was meant to do.

 

[starthaus]

Again, that is everything it was meant to do.

It is good that you are starting to realize the reality.
In what post did you set the problem to only finding the acceleration at the release point?
I asked you this question, why don't you answer it?
Can you solve any of these:

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

You have been given all the tools.

 

[espen180]

It is good that you are starting to realize the reality.
In what post did you set the problem to only finding the acceleration at the release point?
I asked you this question, why don't you answer it?

Posts #50, #64, #66, #68 and #77 all specify that dr/ds=0 was a momentary, inital condition. The fact that the acceleration obtained only holds at that instant is so basic that it shouldn't need mentioning. Aren't you reading the thread?